 Transverse polarization constraint in an angular spectrum

The transverse polarization constraint in an angular spectrum requires each source-free electromagnetic plane wave component to have electric field orthogonal to its wavevector in the algebraic sense:For an isotropic homogeneous medium with constant permittivity, this follows by inserting a Fourier component into $\nabla \cdot \mathbf{D}=0$. It restricts the vector fields that can accompany a scalar angular spectrum method calculation.^[ref]

With time dependence $e^{-i\omega t}$ and $\mathbf{k}=(k_x,k_y,k_z)$, the relation becomesWhen $k_z\ne 0$, the longitudinal field is therefore ${\tilde{E}}_z=-(k_x{\tilde{E}}_x+k_y{\tilde{E}}_y)/k_z$. It is not an arbitrary third scalar image that can be propagated independently of the first two.

For an original propagating angular-spectrum component with $\mathbf{k}=k(0.6,0,0.8)$, choosing ${\tilde{E}}_x=1$ and ${\tilde{E}}_y=0$ forces ${\tilde{E}}_z=-0.75$. Polarization along $y$ instead gives a purely transverse-to-axis field for this propagation plane. Thus electromagnetic transversality relative to $\mathbf{k}$ does not require zero field along the laboratory $z$ axis.

For an evanescent angular-spectrum component with $\mathbf{k}=k(1.25,0,0.75i)$, the same choice ${\tilde{E}}_x=1$ gives ${\tilde{E}}_z=(5/3)i$. The dot product is not a Hermitian inner product: there is no complex conjugation of $\mathbf{k}$ in the Fourier-transformed divergence equation. The imaginary longitudinal amplitude represents a phase relationship, not permission to discard the constraint.

At grazing propagation, $k_z=0$, division by $k_z$ fails and the original equation must be used. Boundary conditions, material response and the corresponding magnetic field remain necessary to reconstruct a physical solution. In the strongly paraxial approximation, longitudinal components may be small, but that is an approximation derived from the allowed spectrum rather than a replacement for the vector condition.